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Difference between revisions of "1967 IMO Problems/Problem 4"

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<math>\alpha = \arctan \frac{b_0 \cos B \sin C + c_0 \sin B \cos (A_0 - C)}
 
<math>\alpha = \arctan \frac{b_0 \cos B \sin C + c_0 \sin B \cos (A_0 - C)}
 
{b_0 \sin B \sin C + c_0 \sin B \sin (A_0 - C)}</math>
 
{b_0 \sin B \sin C + c_0 \sin B \sin (A_0 - C)}</math>
 +
 +
It is easy to verify that this value is valid, and it is indeed
 +
a point of maximum for <math>f(\alpha)</math>.
 +
 +
Now to answer the "and construct it" part of the problem, we
 +
will show that everything we did is constructible, rather than
 +
give a lengthy, boring step by step construction.  Recall that
 +
we already discussed that we can construct <math>\triangle ABC</math> if
 +
we know <math>\alpha</math>.  We just need to show that we can construct
 +
<math>\alpha</math>.  We can construct differences of angles, and given an
 +
angle we can construct two segments whose ratio is the <math>\sin</math>
 +
(or the <math>\cos</math>, or the <math>\tan</math>) of the given angle, and vice-versa.
 +
Given three segments <math>x, y, z</math> we can construct the segment
 +
<math>x \cdot \frac{y}{z}</math>.  Thus, the expression giving <math>\alpha</math>
 +
is constructible.
 +
 +
(Solution by pf02, September 2024)
 +
 +
 +
==(Remarks (added by pf02, September 2024)==
 +
  
  
Line 76: Line 97:
  
  
−
(Solution by pf02, September 2024)
 
  
 
TO BE CONTINUED.  I AM SAVING MID WAY SO AS NOT TO LOSE WORK DONE SO FAR.
 
TO BE CONTINUED.  I AM SAVING MID WAY SO AS NOT TO LOSE WORK DONE SO FAR.

Revision as of 20:00, 3 September 2024

Let $A_0B_0C_0$ and $A_1B_1C_1$ be any two acute-angled triangles. Consider all triangles $ABC$ that are similar to $\triangle A_1B_1C_1$ (so that vertices $A_1$, $B_1$, $C_1$ correspond to vertices $A$, $B$, $C$, respectively) and circumscribed about triangle $A_0B_0C_0$ (where $A_0$ lies on $BC$, $B_0$ on $CA$, and $AC_0$ on $AB$). Of all such possible triangles, determine the one with maximum area, and construct it.


Solution

We construct a point $P$ inside $A_0B_0C_0$ s.t. $\angle X_0PY_0=\pi-\angle X_1Z_1Y_1$, where $X,Y,Z$ are a permutation of $A,B,C$. Now construct the three circles $\mathcal C_A=(B_0PC_0),\mathcal C_B=(C_0PA_0),\mathcal C_C=(A_0PB_0)$. We obtain any of the triangles $ABC$ circumscribed to $A_0B_0C_0$ and similar to $A_1B_1C_1$ by selecting $A$ on $\mathcal C_A$, then taking $B= AB_0\cap \mathcal C_C$, and then $B=CA_0\cap\mathcal C_B$ (a quick angle chase shows that $B,C_0,A$ are also colinear).

We now want to maximize $BC$. Clearly, $PBC$ always has the same shape (i.e. all triangles $PBC$ are similar), so we actually want to maximize $PB$. This happens when $PB$ is the diameter of $\mathcal C_B$. Then $PA_0\perp BC$, so $PC$ will also be the diameter of $\mathcal C_C$. In the same way we show that $PA$ is the diameter of $\mathcal C_A$, so everything is maximized, as we wanted.

This solution was posted and copyrighted by grobber. The thread can be found here: [1]


Solution 2

Since all the triangles $\triangle ABC$ circumscribed to $\triangle A_0B_0C_0$ are similar, the one with maximum area will be the one with maximum sides, or equivalently, the one with maximum side $BC$. So we will try to maximize $BC$.

The plan is to find the value of $\alpha$ which maximizes $BC$.

Prob 1967 4 fig1.png

Note that for any $\alpha$ we can construct the line through $A_0$ which forms the angle $\alpha$ with $A_0C_0$. We can construct points $B, C$ on this line, and lines through these points which form angles $\angle B, \angle C$ with the line, and which pass through $C_0, B_0$ respectively. Since $\triangle A_0B_0C_0, \triangle A_1B_1C_1$ are acute, $A_0$ is between $B, C$ and these lines will meet at a point $A$ such that $B_0$ is between $A, C$ and $C_0$ is between $A, B$.

(More about this later.)

The quantities $a_0, b_0, c_0, \angle B, \angle C$ are given. From this data, $\angle A_0, \angle B_0, \angle C_0, \angle A$ are known and constructible. We will compute $BC$ in terms of $\angle \alpha$ and these quantities. This will be a function in the variable $\alpha$, and we will find the value of $\alpha$ for which this function attains its maximum.

We will start by computing $A_0B$. We will use the law of sines in $\triangle A_0C_0B$. We get $\frac{A_0B}{\sin (\pi - B - \alpha)} = \frac{b_0}{\sin B}$, and a similar equality from $\triangle A_0B_0C$ (for $A_0C$). We obtain

$BC = A_0B + A_0C = \frac{b_0}{\sin B} \sin (B + \alpha) + \frac{c_0}{\sin C} \sin (A_0 - C + \alpha) = f(\alpha)$

We can now proceed in two ways. We could use the formula for linear combination of sine functions with same period but different phase shifts (see https://en.wikipedia.org/wiki/List_of_trigonometric_identities#Linear_combinations or https://mathworld.wolfram.com/HarmonicAdditionTheorem.html , (13)-(23)) or use calculus to find $\alpha$ for which $f(\alpha)$ has its maximum value.

With the first method, we would obtain that $f(\alpha) = D \sin (\alpha + \theta)$ for certain $D$ and $\theta$, and we would choose $\alpha$ such that $\alpha + \theta = \pi/2$. But we will use calculus, as a more mainstream approach. Compute the derivative $f'(\alpha)$ and consider the equation $f'(\alpha) = 0$. Use the formula for $\cos$ of sum of angles, and rearrange terms.

We have

$\cos \alpha \cdot \left [ \frac{b_0}{\sin B} \cdot \cos B + \frac{c_0}{\sin C} \cdot \cos (A_0 - C) \right ]= \sin \alpha \cdot \left [ \frac{b_0}{\sin B} \cdot \sin B + \frac{c_0}{\sin C} \cdot \sin (A_0 - C) \right ]$

Finally,

$\alpha = \arctan \frac{b_0 \cos B \sin C + c_0 \sin B \cos (A_0 - C)} {b_0 \sin B \sin C + c_0 \sin B \sin (A_0 - C)}$

It is easy to verify that this value is valid, and it is indeed a point of maximum for $f(\alpha)$.

Now to answer the "and construct it" part of the problem, we will show that everything we did is constructible, rather than give a lengthy, boring step by step construction. Recall that we already discussed that we can construct $\triangle ABC$ if we know $\alpha$. We just need to show that we can construct $\alpha$. We can construct differences of angles, and given an angle we can construct two segments whose ratio is the $\sin$ (or the $\cos$, or the $\tan$) of the given angle, and vice-versa. Given three segments $x, y, z$ we can construct the segment $x \cdot \frac{y}{z}$. Thus, the expression giving $\alpha$ is constructible.

(Solution by pf02, September 2024)


(Remarks (added by pf02, September 2024)

TO BE CONTINUED. I AM SAVING MID WAY SO AS NOT TO LOSE WORK DONE SO FAR.

See Also

1967 IMO (Problems) • Resources
Preceded by
Problem 3 		1 • 2 • 3 • 4 • 5 • 6 Followed by
Problem 5
All IMO Problems and Solutions
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